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Elenna [48]
3 years ago
13

Bobby tries to push his new big screen TV into the living room. However, Bobby does not push hard enough and cannot move the TV.

List and describe the forces that would be included on the free body diagram of Bobby's TV. Be sure to include the name, direction and brief description for each force. ...?
Physics
1 answer:
Irina18 [472]3 years ago
7 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

<span>Have Bobby as a horizontal force pushing towards/against the tv.
</span><span>Have the force of gravity going downwards from the tv on the floor.
</span><span>Have the force of fric±on between the Foor and the tv

</span>Maybe another force could be bobby's feet pushing from the Foor and his weight (from gravity) bearingdown on his feet. If he didn't weigh more then the tv then he wouldn't be able to put enough pressure<span>on the Foor to create the gripping fric±on force necessary to push the tv</span>
You might be interested in
Help pleasseeee URGENT
Zina [86]

Answer:

The speed of the 8-ball is 2.125 m/s after the collision.

Explanation:

<u>Law Of Conservation Of Linear Momentum</u>

The total momentum of a system of masses is conserved unless an external force is applied. The momentum of a body with mass m and velocity v is calculated as follows:

P=mv

If we have a system of masses, then the total momentum is the sum of all the individual momentums:

P=m_1v_1+m_2v_2+...+m_nv_n

When a collision occurs, the velocities change to v' and the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

In a system of two masses, the law of conservation of linear momentum is simplified to:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

The m1=0.16 Kg 8-ball is initially at rest v1=0. It is hit by an m2=0.17 Kg cue ball that was moving at v2=2 m/s.

After the collision, the cue ball comes to rest v2'=0. It's required to find the final speed v1' after the collision.

The above equation is solved for v1':

\displaystyle v'_1=\frac{m_1v_1+m_2v_2-m_2v'_2}{m_1}

\displaystyle v'_1=\frac{0.16*0+0.17*2-0.17*0}{0.16}

\displaystyle v'_1=\frac{0.34}{0.16}

v'_1=2.125\ m/s

The speed of the 8-ball is 2.125 m/s after the collision.

8 0
3 years ago
You stand at the top of a deep well. To determine the depth, D, of the well you drop a rock from the top of the well and listen
Paladinen [302]

Answer:

(A)

\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s  \right )D+t_t^2v_s^2=0

<em>(B)  D=54.71 m</em>

Explanation:

<u>Free Fall</u>

When a particle is dropped in free air, it starts falling to the ground with an acceleration equal to the gravity. If one wanted to know the height of launching, it can indirectly be measured by the time it takes to reach the ground by the formula

\displaystyle D=\frac{gt^2}{2}

Solving for t

\displaystyle t=\sqrt{\frac{2D}{g}}

If we are taking into consideration the time we can hear the sound it makes when hitting the ground (or water in this case), we must also consider the speed of the sound for the time it takes to reach back our ears. That time can be computed from the basic equation for the speed

\displaystyle t=\frac{D}{v_s}

(A)

The total measured time is the sum of both times and it's given as t_t=3.5\ seconds

\displaystyle t_t=\sqrt{\frac{2D}{g}}+\frac{D}{v_s}

From this equation we'll manage to compute D

First, we isolate the square root

\displaystyle \sqrt{\frac{2D}{g}}=t_t-\frac{D}{v_s}

Let's square both sides

\displaystyle \frac{2D}{g}=t_t^2-2t_t\frac{D}{v_s}+\frac{D^2}{v_s^2}

Multiplying by v_s^2

\displaystyle \frac{2Dv_s^2}{g}=t_t^2v_s^2-2t_tDv_s+D^2

Rearranging and factoring

\boxed{\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s\right )D+t_t^2v_s^2=0}

Now, let's put in numbers:

g=9.8\ m/s^2,\ v_s=345\ m/s,t_t=3.5\ sec

\displaystyle D^2-\left (\frac{2(345)^2}{9.8}+2(3.5)(345)\right )D+(12.25)345^2=0

Computing all the coefficients:

\displaystyle D^2-26,705.82D+1,458,056.25=0

Solving for D, we have two possible solutions:

D=54.71,\ D=26,651.11

The second solution is called "extraneous", since it comes from squaring an equation, which can introduce non-valid (or external) solutions. It's impossible, given the conditions of the problem, that the well could be 26.5 km deep. So we'll keep the only solution as.

<em>D=54.71 m</em>

Let's prove our calculations by computing both times:

\displaystyle t_1=\sqrt{\frac{2(54.71)}{9.8}}=3.34\ sec

\displaystyle t_2=\frac{54.71}{345}=0.16\ sec

We can see their sum is 3.5 seconds, 3.34 of which were taken to reach the bottom of the well, and 0.16 sec took the sound to reach the top.

3 0
3 years ago
A laser beam is incident on two slits with a separation of 0.215 mm, and a screen is placed 5.45 m from the slits. an interferen
vladimir1956 [14]
Use: dsin(θ) = mλ  where d is slit separation, m is fringe order (1 here), and 
θ = 0.183
Now λ = dsin(θ) /m = (0.215e-3)(sin(0.183))/1 = 6.867e-7 or λ = 687.7nm  
(red laser)
4 0
3 years ago
problems like this A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet fi
Ymorist [56]

This question is incomplete the complete question is

A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

Answer:

(a) Xs=0.459m

(b) t=0.984 s

(c) Vc=6.65 m/s

Explanation:

(a) To reach maximum distance

g=-9.8m/s^{2}\\ Vf=0\\v_{b}^{2}=v_{a}^{2}+2gx_{s} \\  x_{s}=\frac{0-(3^{2} )}{-2*9.8}\\ x_{s}=0.459m

(b) For Time

To find t we must find t1 and t2

as

t=t1+t2

For T1

t_{1}=(Vb-Va)/g \\t_{1}=(0-3)/9.8\\t_{1}=0.306s

For T2

x_{l}=Vbt+(1/2)gt_{2}^{2}\\   as\\x_{l}=x_{1}+x_{s}\\x_{l}=1.8+0.459\\x_{l}=2.259\\so\\t_{2}=\frac{2.259*2}{9.8} \\t_{2}=0.6789s

For Total Time

t=t1+t2

t=0.306+0.6789

t=0.984s

(c) To find Vc

Vc=Vb+gt2

Vc=(0)+(9.8)(0.6789)

Vc=6.65 m/s

7 0
3 years ago
"Force" can be defined as a push or a<br> O pull<br> O grab
Nady [450]

Answer:

pull

is your answer please give me some thanks

4 0
2 years ago
Read 2 more answers
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